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Week 1 - Graded Assignment 1

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Course: Jan 2026 - Mathematics I

Topic: Set Theory and Relations | Marks: 1

Question 1

A survey was conducted on pollution of 525 ponds across some cities. It was found that 230 ponds are polluted by fertilisers (F)(F), 245 ponds are polluted by pesticides (P)(P) and 257 ponds are polluted by pharmaceutical products (Ph)(Ph). 100 ponds are polluted by fertilisers and pesticides, 82 ponds are polluted by fertilisers and pharmaceutical products, 77 ponds are polluted by pesticides and pharmaceutical products.
Markdown Image
Define a relation on the set of 525 ponds such that two ponds are related if both are polluted by fertilisers and pharmaceutical products. Which of the following is/are true?
Type: MSQ | Marks: 1 | Topic: Relations
MCQ
Set Theory and Relations
3 Unit Assessment
A
Relation is reflexive.
B
Relation is transitive.
C
Relation is symmetric.
D
This is an equivalence relation.

Topic: Identity Relations and Functions | Marks: 1

Question 2

Consider the relation R={(x,y)โˆฃxโˆ’y=0}โŠ‚Rร—RR= \{(x, y) \mid x-y = 0 \} \subset \mathbb{R} \times \mathbb{R} on the set R\mathbb{R}. Which of the following is/are true?
Type: MSQ | Marks: 1 | Topic: Relations
MCQ
Set Theory and Relations
3 Unit Assessment
A
RR is a transitive relation.
B
RR is a function.
C
RR is not an equivalence relation.
D
RR is a reflexive relation.
E
RR is a symmetric relation.

Topic: Injective (One-One) Functions | Marks: 1

Question 3

Which of the following relations is/are one-one function?
Type: MCQ | Marks: 1 | Topic: Functions
MCQ
Set Theory and Relations
3 Unit Assessment
A
R1={(x,y)โˆฃx,yโˆˆR,x+y>2}R_1 = \{(x, y) \mid x, y \in \mathbb{R}, x+y>2 \}
B
R2={(x,y)โˆฃx,yโˆˆR,x>y}R_2 = \{(x, y) \mid x, y \in \mathbb{R}, x>y \}
C
R3={(x,y)โˆฃx,yโˆˆR,ย x+y=12}R_3 = \{(x,y) \mid x,y \in \mathbb{R},\ x + y = 12\}
D
R4={(x,y)โˆฃx,yโˆˆR,y=x2}R_4 = \{(x, y) \mid x, y \in \mathbb{R}, y=x^2 \}

Topic: Surjective (Onto) and Piecewise Functions | Marks: 1

Question 4

Which of the following is/are true?
Type: MSQ | Marks: 1 | Topic: Functions
MCQ
Set Theory and Relations
3 Unit Assessment
A
Function f:Nโ†’Rf: \mathbb{N} \to \mathbb{R} such that f(x)=x2f(x) = x^2 is not onto.
B
Relation R={(1,1),(1,2),(3,1)}R = \{(1,1),(1,2), (3,1) \} on a set A={1,2,3}A = \{1, 2, 3\} is a function.
C
Function f:Rโ†’Rf: \mathbb{R} \to \mathbb{R} such that f(x)=โˆฃxโˆฃf(x) = |x| is not one-one.
D
Function f:Rโ†’Rf: \mathbb{R} \to \mathbb{R} such that f(x)=(xโˆ’1)2f(x) = (x-1)^2 is one-one and onto.

Topic: Set Cardinality and Regions | Marks: 1

Question 5

Consider the following sets,
A={xโˆฃxโˆˆN}A = \{x \mid x \in\mathbb{N}\}
B={xโˆฃโˆ’5<x<105,xโˆˆR}B = \{x \mid -5< x< 105, x \in \mathbb{R} \}
C={xโˆฃxย isย aย rationalย number,10<xโ‰ค80}C = \{x \mid x~ \text{is a rational number}, 10< x \leq 80 \}
Find the cardinality of the set (Aโˆ’C)โˆฉB(A - C) \cap B.
Type: NAT | Marks: 1 | Topic: Set Theory
MCQ
3 Unit Assessment

Reference-only archive item

The completed export preserved the prompt as an image but not a reusable answer key. The reconstruction below is for study, not scoring.

Study reconstruction
Accepted Answer: 35 Abstract Solution (Strategy):
  • Set Difference (Aโˆ’CA - C): Remove elements from AA that are also in CC. Since AA is N\mathbb{N} and CC contains rationals between 10 and 80, this removes all natural numbers xx such that 10<xโ‰ค8010 < x \leq 80.
  • Intersection with BB: Restrict the remaining set to the range (โˆ’5,105)(-5, 105).
  • Sum the Counts: Count natural numbers in the distinct segments. ^maths-w1-q5-strategy
Procedure:
  • A={1,2,3,...}A = \{1, 2, 3, ...\} (Note: In IITM Maths 1, N\mathbb{N} usually starts from 1, but sometimes 0 is included. Let's verify: โˆฃ(Aโˆ’C)โˆฉBโˆฃ|(A-C)\cap B|).
  • CC removes integers {11,12,...,80}\{11, 12, ..., 80\} from N\mathbb{N}.
  • Remaining in AA: {1,2,...,10,81,82,...}\{1, 2, ..., 10, 81, 82, ...\}.
  • Intersection with BB ((โˆ’5,105)(-5, 105)): {1,2,...,10}โˆช{81,82,...,104}\{1, 2, ..., 10\} \cup \{81, 82, ..., 104\}.
  • Count in {1...10}=10\{1...10\} = 10.
  • Count in {81...104}=104โˆ’81+1=24\{81...104\} = 104 - 81 + 1 = 24.
  • Total = 10+24=3410 + 24 = 34.
  • Wait, Check IITM definition of N\mathbb{N}: If the accepted answer is 35, it implies N={0,1,2,...}\mathbb{N} = \{0, 1, 2, ...\}.
  • Then {0,1,...,10}\{0, 1, ..., 10\} has 11 elements. 11+24=3511 + 24 = 35.

Topic: Principle of Inclusion-Exclusion (PIE) | Marks: 1

Question 6

A survey was conducted on pollution of 525 ponds across some cities. It was found that 230 ponds are polluted by fertilisers (F)(F), 245 ponds are polluted by pesticides (P)(P) and 257 ponds are polluted by pharmaceutical products (Ph)(Ph). 100 ponds are polluted by fertilisers and pesticides, 82 ponds are polluted by fertilisers and pharmaceutical products, 77 ponds are polluted by pesticides and pharmaceutical products.
Markdown Image
Find the number of ponds polluted by all types of water contaminants.
Type: NAT | Marks: 1 | Topic: Set Theory
MCQ
3 Unit Assessment

Reference-only archive item

The completed export preserved the prompt as an image but not a reusable answer key. The reconstruction below is for study, not scoring.

Study reconstruction
Accepted Answer: 52 Abstract Solution (Strategy):
  • Principle of Inclusion-Exclusion (PIE): Use the formula for 3 sets: โˆฃAโˆชBโˆชCโˆฃ=โˆฃAโˆฃ+โˆฃBโˆฃ+โˆฃCโˆฃโˆ’(โˆฃAโˆฉBโˆฃ+โˆฃBโˆฉCโˆฃ+โˆฃCโˆฉAโˆฃ)+โˆฃAโˆฉBโˆฉCโˆฃ|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |C \cap A|) + |A \cap B \cap C|.
  • Plug and Chug: Assuming all ponds are polluted by at least one (since no "none" is mentioned and total is 525). ^maths-w1-q6-strategy
Procedure: โˆฃFโˆชPโˆชPhโˆฃ=โˆฃFโˆฃ+โˆฃPโˆฃ+โˆฃPhโˆฃโˆ’(โˆฃFโˆฉPโˆฃ+โˆฃPโˆฉPhโˆฃ+โˆฃFโˆฉPhโˆฃ)+โˆฃFโˆฉPโˆฉPhโˆฃ|F \cup P \cup Ph| = |F| + |P| + |Ph| - (|F \cap P| + |P \cap Ph| + |F \cap Ph|) + |F \cap P \cap Ph| 525=230+245+257โˆ’(100+77+82)+X525 = 230 + 245 + 257 - (100 + 77 + 82) + X 525=732โˆ’259+X525 = 732 - 259 + X 525=473+X525 = 473 + X X=525โˆ’473=52X = 525 - 473 = 52.

Topic: Modeling Family Trees as Functions | Marks: 1

Question 7

Mahesh has four sons (Shubh, Rabi, Mahendra, and Rajat). Shubh has two sons (Yashubh and Navrtna). Rabi has two sons named Rathi and Rakesh.
Let RR be "are cousins" and SS be "is son of". f=Sโˆฉ(Pร—Q)f = S \cap (P \times Q).
Which of the following options are correct?
Markdown Image
Type: MSQ | Marks: 1 | Topic: Relations & Functions
MCQ
Set Theory and Relations
3 Unit Assessment
A
f:Pโ†’Qf: P \to Q is a function, where P={Yashubh,ย Navrtna,ย Rathi,ย Rakesh,ย Mahesh}P = \{\text{Yashubh, Navrtna, Rathi, Rakesh, Mahesh}\} and Q={ย Shubh,ย Rabi,ย Mahendra,ย Rajat}Q=\{\text{ Shubh, Rabi, Mahendra, Rajat}\}.
B
If f:Pโ†’Qf: P \to Q is a function, where P={ย Yashubh,ย Navrtna,ย Rathi,ย Rakesh}P=\{\text{ Yashubh, Navrtna, Rathi, Rakesh}\} and Q={Shubh,ย Rabi,ย Mahendra,ย Rajat}Q=\{\text{Shubh, Rabi, Mahendra, Rajat}\}, then ff is one-one.
C
If f:Pโ†’Qf: P \to Q is a function, where P={Yashubh,ย Navrtna,ย Rathi,ย Rakesh}P=\{\text{Yashubh, Navrtna, Rathi, Rakesh}\} and Q={ย Shubh,ย Rabi}Q=\{ \text{ Shubh, Rabi}\}, then ff is onto.
D
If f:Pโ†’Qf: P \to Q is a function, where P={Yashubh,ย Rathi}P=\{\text{Yashubh, Rathi}\} and Q={Shubh,ย Rabi}Q=\{\text{Shubh, Rabi}\}, then ff is bijective.

Topic: Domains and Codomains | Marks: 1

Question 8

Consider the table of materials and dielectric constants.
MaterialDielectric constant
Air1
Vaccum2
Paper3
Glass8
Nerve membrane7
Silicon13
Let ff be a function from the set of materials to the set of dielectric constants. Which of the following is true?
Type: MCQ | Marks: 1 | Topic: Functions
MCQ
Set Theory and Relations
3 Unit Assessment
A
ff is neither one to one nor onto.
B
ff is one to one but not onto.
C
ff is onto but not one to one.
D
ff is bijective.

Topic: Symmetric Difference of Sets | Marks: 1

Question 9

Consider the following sets.
  • A={xโˆˆNโˆฃxย modย 2=0ย andย 1โ‰คxโ‰ค10}A = \lbrace x\in \mathbb{N}\mid x \text{ mod } 2 = 0 \text { and } 1\leq x \leq 10 \rbrace
  • B={xโˆˆNโˆฃxย modย 5=0ย andย 6โ‰คxโ‰ค25}B = \lbrace x\in \mathbb{N}\mid x \text{ mod } 5 = 0 \text { and } 6\leq x \leq 25 \rbrace
  • C={xโˆˆNโˆฃxย modย 7=0ย andย 7โ‰คxโ‰ค29}C = \lbrace x\in \mathbb{N}\mid x \text{ mod } 7 = 0 \text { and } 7\leq x \leq 29 \rbrace
What is the cardinality of (Aโˆ–(BโˆชC))โˆช(Bโˆ–(CโˆชA))โˆช(Cโˆ–(BโˆชA))\big (A\setminus (B\cup C) \big) \cup \big (B\setminus (C\cup A)\big )\cup \big (C\setminus (B\cup A)\big )
Type: NAT | Marks: 1 | Topic: Set Theory
MCQ
3 Unit Assessment

Reference-only archive item

The completed export preserved the prompt as an image but not a reusable answer key. The reconstruction below is for study, not scoring.

Study reconstruction
Accepted Answer: 11 Abstract Solution (Strategy):
  • List all elements: Explicitly write down the elements of A,B,CA, B, C.
  • Apply Set Difference: For each set, remove elements found in the other two.
  • Symmetric Difference logic: This expression represents elements belonging to exactly one of the three sets.
Procedure:
  • A={2,4,6,8,10}A = \{2, 4, 6, 8, 10\}
  • B={10,15,20,25}B = \{10, 15, 20, 25\}
  • C={7,14,21,28}C = \{7, 14, 21, 28\}
  • Excluded by overlapping:A \cap B = \{10\}$$B \cap C = \emptyset$$A \cap C = \emptyset
  • Single membership sets:Aโˆ–(BโˆชC)={2,4,6,8}A\setminus(B \cup C) = \{2, 4, 6, 8\} (4 elements)Bโˆ–(CโˆชA)={15,20,25}B\setminus(C \cup A) = \{15, 20, 25\} (3 elements)Cโˆ–(BโˆชA)={7,14,21,28}C\setminus(B \cup A) = \{7, 14, 21, 28\} (4 elements)
  • Total cardinality = 4+3+4=114 + 3 + 4 = 11.

Topic: Multi-set Overlap Word Problems | Marks: 1

Question 10

In 180 people survey: 95 watched Dabangg, 100 watched Avatar, 100 watched RRR. Intersections: Dabangg โˆฉ\cap Avatar = 50, Avatar โˆฉ\cap RRR = 40, Dabangg โˆฉ\cap RRR = 55. If all 180 people watched at least one movie, find the number of people who watched only RRR and Avatar but not Dabangg.
Type: NAT | Marks: 1 | Topic: Set Theory
MCQ
3 Unit Assessment

Reference-only archive item

The completed export preserved the prompt as an image but not a reusable answer key. The reconstruction below is for study, not scoring.

Study reconstruction
Accepted Answer: 10 Abstract Solution (Strategy):
  • Find โˆฃallย 3โˆฃ|\text{all 3}|: Use PIE (180=โˆ‘โˆฃSiโˆฃโˆ’โˆ‘โˆฃSiโˆฉSjโˆฃ+โˆฃS1โˆฉS2โˆฉS3โˆฃ180 = \sum |S_i| - \sum |S_i \cap S_j| + |S_1 \cap S_2 \cap S_3|).
  • Isolated Overlap: "Only A and R" means (AโˆฉR)โˆ–D(A \cap R) \setminus D.
  • Formula: โˆฃOnlyย AโˆฉRโˆฃ=โˆฃAโˆฉRโˆฃโˆ’โˆฃAโˆฉRโˆฉDโˆฃ|\text{Only } A \cap R| = |A \cap R| - |A \cap R \cap D|.
Procedure: 180=95+100+100โˆ’(50+40+55)+X180 = 95 + 100 + 100 - (50 + 40 + 55) + X 180=295โˆ’145+Xโ€…โ€ŠโŸนโ€…โ€Š180=150+Xโ€…โ€ŠโŸนโ€…โ€ŠX=30180 = 295 - 145 + X \implies 180 = 150 + X \implies X = 30.
  • People who watched all 3 movies = 30.
  • People who watched Avatar and RRR = 40.
  • People who watched only Avatar and RRR = 40โˆ’30=1040 - 30 = 10.

Question 11

Family Tree relations RR ("are cousins") and SS ("is son of"). Mahesh has four sons (Shubh, Rabi, Mahendra, and Rajat). Shubh has two sons (Yashubh and Navrtna). Rabi has two sons named Rathi and Rakesh.
Which of the following options are correct?
Type: MSQ | Marks: 1 | Topic: Relations
MCQ
Set Theory and Relations
3 Unit Assessment
A
RR is an equivalence relation.
B
(Rathi, Navrtna) โˆˆR\in R
C
SS is an equivalence relation.
D
(Mahesh, Rajat) โˆˆS\in S.
E
(Yashubh, Rathi) โˆˆR\in R and (Rathi, Yashubh) โˆˆR\in R.
F
SS is symmetric relation.

Question 12

Family Tree: Let mm be the number of elements in RR (cousins) and nn be the number of elements in SS (son of). Find m+nm+n. (Consider the set consisting of all mentioned people).
Type: NAT | Marks: 1 | Topic: Relations
MCQ
3 Unit Assessment

Reference-only archive item

The completed export preserved the prompt as an image but not a reusable answer key. The reconstruction below is for study, not scoring.

Study reconstruction
Accepted Answer: 16 Abstract Solution (Strategy):
  • List SS pairs: Count every (child, parent) edge.
  • List RR pairs: Count every (cousin1, cousin2) pair. Note: Relations are sets of ordered pairs (x,y)(x,y).
  • Cousin Calculation: If group 1 has xx people and group 2 has yy people, the number of cousin pairs (g1,g2)(g1, g2) and (g2,g1)(g2, g1) is 2xy2xy.
Procedure:
  • SS (Son of):(Shubh, Mahesh), (Rabi, Mahesh), (Mahendra, Mahesh), (Rajat, Mahesh) โ†’\to 4(Yashubh, Shubh), (Navrtna, Shubh) โ†’\to 2(Rathi, Rabi), (Rakesh, Rabi) โ†’\to 2**n=8n = 8**.
  • RR (Cousins):Group 1 (Shubh's kids): {Yashubh, Navrtna}Group 2 (Rabi's kids): {Rathi, Rakesh}Pairs: (Y, Rathi), (Y, Rakesh), (N, Rathi), (N, Rakesh) and inverses.Count = 2ร—2ร—2=82 \times 2 \times 2 = 8.m=8m = 8.
  • Total = m+n=16m + n = 16.

Question 13

Let A=QA = \mathbb{Q} and B=ZB = \mathbb{Z}. Consider the function f:Aโ†’Bf: A \to B defined as f(p/q)=pโˆ’qf(p/q) = p - q where gcd(p,q)=1gcd(p, q) = 1 and q>0q > 0. Which of the following is true?
Type: MCQ | Marks: 1 | Topic: Functions
MCQ
Set Theory and Relations
3 Unit Assessment
A
ff is one to one but not onto.
B
ff is neither one to one nor onto.
C
ff is onto but not one to one.
D
ff is bijective.

Question 14

Match the following sets: i.ย Finitei. \text{ Finite} ii.ย Infiniteii. \text{ Infinite} iii.ย Singletoniii. \text{ Singleton} iv.ย Emptyiv. \text{ Empty}
Markdown Image

(A) {xโˆฃx2=25,xโˆˆN}\{x \mid x^2 = 25, x \in \mathbb{N} \} (B) {xโˆฃx2+1=0,xโˆˆR}\{x \mid x^2 + 1 = 0, x \in \mathbb{R} \} (C) {xโˆฃx2โˆ’xโˆ’6=0,xโˆˆZ}\{x \mid x^2 - x - 6 = 0, x \in \mathbb{Z} \} (D) {xโˆฃx+2>0,xโˆˆR}\{x \mid x+2 > 0, x \in \mathbb{R} \}
Type: MCQ | Marks: 1 | Topic: Set Theory
MCQ
Set Theory and Relations
3 Unit Assessment
A
Aโ†’iii,Bโ†’iv,Cโ†’i,Dโ†’iiA \to iii, B \to iv, C \to i, D \to ii
B
Aโ†’iii,Bโ†’i,Cโ†’iv,Dโ†’iiA \to iii, B \to i, C \to iv, D \to ii
C
Aโ†’iii,Bโ†’iv,Cโ†’i,Dโ†’iiA \to iii, B \to iv, C \to i, D \to ii (Wait, same as A)
D
Aโ†’iii,Bโ†’ii,Cโ†’iv,Dโ†’iA \to iii, B \to ii, C \to iv, D \to i

Question 15

If f1f_1 and f2f_2 are two functions with domains D1D_1 and D2D_2 respectively, then the domain of f1+f2f_1 + f_2 is:
Type: MCQ | Marks: 1 | Topic: Functions
MCQ
Set Theory and Relations
3 Unit Assessment
A
D1โˆชD2D_1 \cup D_2
B
D1โˆ–D2D_1 \setminus D_2
C
D2โˆ–D1D_2 \setminus D_1
D
D1โˆฉD2D_1 \cap D_2

Question 16

Let f,g:Rโ†’Rf, g: \mathbb{R} \to \mathbb{R} be defined as f(x)={xxโ‰ค0ย 0x>0f(x) = \begin{cases} x & x \leq 0 \ 0 & x > 0 \end{cases} g(x)={0xโ‰ค0ย xx>0g(x) = \begin{cases} 0 & x \leq 0 \ x & x > 0 \end{cases} Then f+gf + g is:
Type: MSQ | Marks: 1 | Topic: Functions
MCQ
Set Theory and Relations
3 Unit Assessment
A
ff is one-to-one but not onto.
B
ff is neither.
C
gg is onto but not one-to-one.
D
f+gf+g is bijective.

Question 17

Consider the relations R1={(a,b)โˆˆNร—Nโˆฃb=a+1}R_1 = \{(a, b) \in \mathbb{N} \times \mathbb{N} \mid b = a + 1 \} R2={(a,b)โˆˆNร—Nโˆฃbโ‰ฅa}R_2 = \{(a, b) \in \mathbb{N} \times \mathbb{N} \mid b \geq a \} Which of the following is true?
Type: MSQ | Marks: 1 | Topic: Relations
MCQ
Set Theory and Relations
3 Unit Assessment
A
R1R_1 is reflexive.
B
R2R_2 is symmetric.
C
R1R_1 is neither symmetric nor transitive.
D
R2R_2 is both reflexive and transitive.

Question 18

Let A={3,6,9,...}A = \{3, 6, 9, ... \} (set of positive multiples of 3). Let f:Nโ†’Af: \mathbb{N} \to A be defined as f(n)=6nf(n) = 6n. Is ff bijective?
Type: MSQ | Marks: 1 | Topic: Functions
MCQ
Set Theory and Relations
3 Unit Assessment
A
ff is one-to-one but not onto.
B
ff is neither.
C
ff is onto but not one-to-one.
D
ff is bijective.


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